Question

Evaluate ${\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\tan x}}$.

• A. $\frac{\pi }{12}$
• B. $\frac{7\pi }{12}$
• C. $\frac{5\pi }{12}$
• D. None of these

Answer 1

The correct option is A $\frac{\pi }{12}$

$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dx}{1+\sqrt{\tan x}}$

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dx}{1+\sqrt{\tan (\frac{\pi }{3}+\frac{\pi }{6}-x)}}$

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dx}{1+\sqrt{\cot x}}$

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{\tan x}dx}{1+\sqrt{\tan x}}$

Adding another I in the above equantion, we get

$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{\tan x}dx}{1+\sqrt{\tan x}}+{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dx}{1+\sqrt{\tan x}}$

$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}dx$

$2I=\frac{\pi }{6}$

$I=\frac{\pi }{12}$